# A 100.0 mL sample of a solution that is 0.125 M in HClO and 0.130 M in NaClO has added to it 2.00 mL of 0.100 M HCl. The Ka for HClO is 2.9 x 10-8. What is the pH of the resulting solution?

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To start, I found the moles of each compound, then I added the moles of HClO and HCl to get my total moles of acid. Finally, I plugged it into the Henderson-Hasselbach equation.

I found that 0.0002 moles of HCl was added to a solution that contained 0.0125 moles of HClO and 0.013 moles of NaClO

So...

-log(2.9 -log(#2.9*10^-8# ) + log(#0.013/0.0127# )

pH=7.55

Does this seem right? I wasn't sure how to approach this problem and I am unsure about the answer I got.

Thanks for any help!

To start, I found the moles of each compound, then I added the moles of HClO and HCl to get my total moles of acid. Finally, I plugged it into the Henderson-Hasselbach equation.

I found that 0.0002 moles of HCl was added to a solution that contained 0.0125 moles of HClO and 0.013 moles of NaClO

So...

-log(2.9 -log(

pH=7.55

Does this seem right? I wasn't sure how to approach this problem and I am unsure about the answer I got.

Thanks for any help!

##### 1 Answer

#### Explanation:

Start by finding the quantity of each species in the

#n("HCl") = c * V = 2.00 xx 10^(-4) color(white)(l) mol# #n("HClO") = c * V = 1.25 xx 10^(-2) color(white)(l) mol# # n("NaClO") = c * V = 1.30 xx 10^(-2) color(white)(l) mol#

Hydrochloric acid

The rest of

The hydrolysis of aqueous **"RICE"** table can help with the derivation of the equilibrium proton concentration due to this reversible process.

**R**

**I**

**C**

**E**

(all values in this table are in

Where **RICE** table,

Solving this equation for

The final solution has a volume of

and therefore